# mathShow that sin (2a+b)=sin a if cos (a+b)=1.

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The problem provides the information that `cos(a + b) = 1` , hence, ` a + b = cos^(-1)1 => a + b = 0` , thus `a = -b` .

You need to test if `sin (2a+b) = sin a` , under the given condition, `cos(a + b) = 1` , such that:

`sin (2a+b) = sin a => 2a + b = a => 2a - a = -b => a = -b` valid

**Hence, testing the statement `sin (2a+b) = sin a` , using the condition `cos(a + b) = 1` , yields that `sin (2a+b) = sin a` holds.**

We'll expand cosine of the sum:

cos (a+b) = cos a*cos b - sin a*sin b

From enunciation, we know that:

cos a*cos b - sin a*sin b = 1

cos a*cos b = 1 + sin a*sin b (1)

Now, we'll expand the function sin (2a+b):

sin (2a+b) = sin 2a*cos b + sin b*cos 2a

We'll re-write the factor sin 2a:

sin 2a = sin(a+a) = 2sin a*cos a

We'll re-write the factor cos 2a:

cos 2a = cos (a+a) = 1 - 2(sin a)^2

We'll re-write the sum:

sin (2a+b) = 2sin a*cos a*cos b + sin b*[1 - 2(sin a)^2]

We'll substitute the product cos a*cos b by (1):

sin (2a+b) = 2sin a*(1 + sin a*sin b) + sin b*[1 - 2(sin a)^2]

We'll remove the brackets:

sin (2a+b) = 2sin a + 2(sin a)^2*sin b + sin b - 2(sin a)^2*sin b

sin (2a+b) = 2sin a + sin b